1. ## need help gr12

can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx

when a square is inscribed in a circle, the length of the diagnol of the square =the diameter of the circle
2^(1/2)side of square = diagonal of square
area of square = 1/2 * diagonal * diagonal
S1 = 2 (diagonal of S1 =2)
when a circle is inscribed in a square, length of square = diameter of circle
diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)
S2 = 1/2 * 2 = 1
Sn = 2*[(1/2)^(n-1)]

3. Originally Posted by kansai
can any one do 3,4,5,6,7,8??????
2)
You know that,
$t_1=-2$
$t_2=1$
And that,
$t_3=2t_2-t_1$
thus,
$t_3=2(1)-(-2)=2+2=4$
And that,
$t_4=2t_3-t_2$
thus,
$t_4=2(4)-1=8-1=7$
And that,
$t_5=2t_4-t_3$
thus,
$t_5=2(7)-1=13$

4. Hello, kansai!

Here's #5 . . .

5. For the matrices: $A = \begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix}$ and $B = \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$

(a) State the value of: $\text{(i) }a_{12}\qquad\text{(ii) }a_{22}\qquad\text{(iii) }a_{31}$

(b) Calculate: $\text{(i) }A - 3B\qquad\text{(ii) }2A + B\qquad\text{(iii) }A'$

$\text{(a) }\;a_{12}$ means the element in the $1^{st}$ row and the $2^{nd}$column. .Hence: $a_{12} = 5$
. . . .Similarly: $a_{22} = 0,\;a_{31} = 2$

The elements of matrix $B$ would have been written: $b_{12},\;b_{22},\;b_{31}$

$\text{(b)(i) }A - 3B \:=\:\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - 3\cdot\begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$

. . . $=$ $\;\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - \begin{bmatrix}6&0&-3\\3&0&3\\9&-6&0\end{bmatrix} \;=\;\begin{bmatrix}-5&5&3\\0&0&1\\-7&6&-1\end{bmatrix}$

$\text{(b)(ii) }2A + B \;= \;2\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$

. . . $=\;\begin{bmatrix}2&10&0\\6&0&8\\4&0&-2\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix} \;= \;\begin{bmatrix}4&10&-1\\7&0&9\\7&-2&-2\end{bmatrix}$

$\text{(b)(iii) }A'$ . . . I assume this means the inverse of $A.$

The steps are long and messy . . . so I'll omit them.

. . . $A'\;=\;\frac{1}{55}\begin{bmatrix}0&5&20\\11&-1&-4\\0&10&-15\end{bmatrix}$

5. Hello, kansai!

Here's #3 . . .

For the map given on the right
(a) Represent the map with a network.
(b) Find the degree of each vertex.
(c) State whether the network is traceable and explain the reason.
Code:
(a)

A - - - - - B
|         / | \
|       /   |  \
|     C     |   \
|   /       |    \
| /         |     \
D - - - - - E - - -F
(b) A: degree 2
. . .B: degree 4
. . .C: degree 2
. . .D: degree 3
. . .E: degree 3
. . .F: degree 2

(c) A network is traceable if it has at most two vertices of odd degree.

This network has two odd vertices (D and E) and hence is traceable.

6. Hello, kansai!

I don't "trust" #4 . . . It has some disturbing errors and the meaning is unclear.
. . Can you provide the original wording?

4. The NYS school has four committees.
Each of these committees meet once a month.

Membership on these committees is as follows:
. . Committee A: Anna, John, Steven, Tom
. . Committee B: Debra, Jane John
. . Committee C: Annie, Debra, Larry, Tom
. . Committee D: Bobby, Debra, Grace, Mary, Steven
. . Committee E: Anna, Larry, Tom.

(a) Draw a network to illustrate the connectivity among the committees. .?

(b) Design a schedule with minimum timeslots for the committee meeting without conflicts. .??
A petty point: I assume it refers to The "NYS school system".

I would assume that Annie is the same person as Anna, that someone mistyped the names.
. . But do I dare make that assumption?

The most glaring errors is that there are five committees.

(a) Connectivity is a term used in network theory, but not like this.
. . I assume they mean "common memberships" ?

(b) Minimum timeslots is a very sloppy term.

Does it mean "the shortest meetings"?
. . How about: "I call the meeting to order. .I will entertain a motion to adjourn . . ."

Does it mean "the shortest time-span" for the four (five?) meetings?
. .Schedule them consecutively with a one-minute break in between
. . (to allow the ten people to switch seats).

It is obvious that no two committees can meet at the same time.
. . So what is the question?

If I am correct about part (a), this should suffice:
Code:
                  B
/ | \
/   |   \
/     |     \
A - - - + - - - D
/  \     |     /
/     \   |   /
/        \ | /
E - - - - - C

7. ## URGENT can any one do question 6,7,8?

URGENT can any one do question 6,7,8?? and plz show ur steps and explation..............thnx

8. Originally Posted by malaygoel
when a square is inscribed in a circle, the length of the diagnol of the square =the diameter of the circle
area of square = 1/2 * diagonal * diagonal
S1 = 1/2
when a circle is inscribed in a square, length of square = diameter of circle
diameter of O2= (1/2)^(1/2)
S2 = 1/2 * 1/2 = 1/4
Sn = (1/2)^n
The questions to this test were already answered in General high school math-need help gr12

9. ## Question 1

From my understanding of this question it works out like this.

the diagnol of the square (d) = the diameter of the circle (d), so the area =

$s$ = the length of one of the square's side

$d=\sqrt{ s^2 +s^2}$
$d^2=s^2 +s^2$
$d^2=2s^2$
$\frac{d^2}{2}=s^2$ since s^2 is the area of the square this is the answer for $S_1$.
Now the next circle has the diameter equal to the height of the square, so...
$\sqrt{\frac{d^2}{2}}=s=d_2$
and now you apply the first formula to find the area of $S_2$
$\sqrt{\frac{d^2}{2}}=\sqrt{s^2 +s^2}$
$\sqrt{\left(\frac{d^2}{2}\right)^2}=s^2 +s^2$
$\frac{d^2}{2}=2s^2$
$\frac{\frac{d^2}{2}}{2}=s^2$
$\frac{d^2}{4}=s^2$
$\frac{d^2}{4}=s^2$

This shows us that the area of $S_n = \frac{d^2}{2^n}$

Considering me and malaygoel got different answers, could a third party check my work?

10. Originally Posted by Quick
From my understanding of this question it works out like this.

the diagnol of the square (d) = the diameter of the circle (d), so the area =

$s$ = the length of one of the square's side

$d=\sqrt{ s^2 +s^2}$
$d=s +s$
$d=2s$
how did you get d=s+s

11. how did you get d=s+s
I made a mistake, which I corrected on the post, but I suppose it doesn't matter since he was asking for #6 instead of #1!

12. Originally Posted by Quick

Considering me and malaygoel got different answers, could a third party check my work?
I have corrected my answer in the post. Could you please check it?

13. diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)
it might be that I'm tired, buy I can't seem to think where you got the 2^(-1/2) from. However, you can see our equations differ,
mine= $\frac{d^2}{2^n}$ yours= $2\left(\frac{1}{2^{(n-1)}}\right)$

However, there is a certain value of d that would make our equations the same.

$\frac{d^2}{2^n}=2\left(\frac{1}{2^{(n-1)}}\right)$
$\frac{d^2}{2^n}=\frac{2}{2^{(n-1)}}$
$\frac{2^2}{2^n}=\frac{2}{2^{(n-1)}}$
$\frac{4}{2^n}=\frac{2}{2^{(n-1)}}$
$\frac{2}{2^{(n-1)}}=\frac{2}{2^{(n-1)}}$

but why would you say the diameter is 2?

14. friend got these answer for #1

plz tell me is it correct?

area of 1st square = 2
area of 2nd square= 1
area of 3rd square= 2^-1
area of Sn square= 2^2-n

15. assuming that the diameter of the first circle is 2, then yes.

$\frac{d^2}{2^n}$
$\frac{2^2}{2^n}$
$\frac{4}{2^n}$
$\frac{4}{2^n}$
$\frac{1}{2^{(n-2)}}$
$2^{(-1)(n-2)}$
$2^{(-n+2)}$
$2^{(2-n)}$

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